Calculate the binding energy per nucleon of `._20Ca^(40)` nucleus. Given m `._20Ca^(40)=39.962589u` , `M_(p) =1.007825u` and `M_(n)=1.008665 u` and Take `1u=931MeV`.

Explanation

To calculate the binding energy per nucleon of the \( _{20}^{40}\text{Ca} \) nucleus, we will follow these steps:

### Step 1: Identify the Number of Protons and Neutrons
The atomic number \( Z \) of Calcium is 20, which means it has 20 protons. The mass number \( A \) is 40, which means the total number of nucleons (protons + neutrons) is 40. Therefore, the number of neutrons \( N \) can be calculated as:
\[ N = A - Z = 40 - 20 = 20 \]

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